Question

The lifetime of a lightbulb can be modeled with an exponential random variable with an expected lifetime of 1000 days. (a) Find the probability that the lightbulb will function for more than 2000 days. (b) Find the probability that the lightbulb will function for more than 2000 days, given that it is still functional after 500 days

Answer 1

To find the probability that a lightbulb will function for more than 2000 days, we need to use the exponential distribution and calculate the integral of the probability density function (PDF). Similarly, to find the probability that a lightbulb will function for more than 2000 days, given that it is still functional after 500 days, we need to calculate a conditional probability using the PDF.

Step-by-step explanation:

To solve this problem, we can use the exponential distribution. The exponential distribution can be described by the formula f(x) = λ * e^(-λx), where λ is the decay parameter and x is the time.

(a) Probability that the lightbulb will function for more than 2000 days:

Since the expected lifetime of the lightbulb is 1000 days, we can find the value of λ using the formula λ = 1 / expected lifetime. In this case, λ = 1 / 1000 = 0.001.

To find the probability that the lightbulb will function for more than 2000 days, we need to calculate the integral of the probability density function (PDF) from 2000 to infinity. Since the exponential distribution is continuous and non-negative, the integral represents the probability.

The integral can be evaluated as follows:

1. Calculate the integral of the PDF from 0 to 2000 to find the probability that the lightbulb will function for less than or equal to 2000 days. This can be done using the formula P(T <= t) = 1 - e^(-λt).
2. Subtract the result from 1 to find the probability that the lightbulb will function for more than 2000 days. This can be done using the formula P(T > t) = 1 - P(T <= t).

(b) Probability that the lightbulb will function for more than 2000 days, given that it is still functional after 500 days:

To find this conditional probability, we need to calculate the probability that the lightbulb will function for more than 2000 days and it is still functional after 500 days, divided by the probability that it is still functional after 500 days. This can be done using the formula P(T > 2000 | T > 500) = P(T > 2000 and T > 500) / P(T > 500).

To calculate P(T > 2000 and T > 500), we can subtract the probability that the lightbulb will function for less than or equal to 500 days from the probability that it will function for less than or equal to 2000 days. This can be done using the formula P(T <= t) = 1 - e^(-λt).

To calculate P(T > 500), we can subtract the probability that the lightbulb will function for less than or equal to 500 days from 1.

Answer 2

a) 0.13533

b) 0.36788

Step-by-step explanation:

Let X be the random variable that measures the lifetime of a bulb.

If the random variable X is exponentially distributed and X has an average value of 1000 days, then its probability density function is

`\bf f(x)=(1)/(1000)e^(-x/1000)\;(x\geq 0)`

and its cumulative distribution function (CDF) is

`\bf P(X\leq t)=\int_(0)^(t) f(x)dx=1-e^(-t/1000)`

(a) Find the probability that the light bulb will function for more than 2000 days

`\large P(X>2000)=1-P(X\leq 2000)=\\\\=1-(1-e^(-2000/1000))=e^(-2)=0.135335`

(b) Find the probability that the light bulb will function for more than 2000 days, given that it is still functional after 500 days

Let A and B be the events,

A = “The bulb will last at least 2000 days”

B = “The bulb will last at least 500 days”

We want to find P(A | B).

By definition P(A | B) = P(A∩B)P(B)

but B⊂A, so A∩B = B and

`\large P(A | B) = P(B)P(B) = (P(B))^2`

We have

`\large P(B)=P(X \geq 500)=1-P(X\leq 500)=1-(1-e^(-500/1000))=e^(-0.5)=0.60653`

hence,

`\large P(A | B)=(P(B))^2=(0.60653)^2=0.36788`